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What's considered an interaction in Quantum Mechanics?

Physics Asked by strategaD on February 23, 2021

Clearly I’m not an expert of QM but recently I came across the fact that particles do not have intrinsically a specific location until they interact (very ambigous term to me) with something else and so they wave function collapses into a specific location. My question is: what’s considered an interaction? Doesn’t particles interact with each other trough gravity and electromagnetism all the time? The effect of gravity is an interaction mediated by the deformation of space-time and so may be excluded but to me EM should make the wave function collapse every instant.

3 Answers

In the quantum mechanical frame, all matter is ultimately composed from elementary particles , at the moment given in the table of particles of the standardm model of particle physics. this is a quantum mechanical model in the framework of special relativity.

This means that Lorentz transformations dictate the kinematics of four vectors, for spacetime $(t,x,y,z)$ and for energy-momentum $(E,p_x.p_y,p_z)$.

An interaction involves two incoming particle four vectors, on mass shell , i.e. the "length" of the energy momentum four vector is on mass shell, which scatter off each other, ending with two or more particle on mass shell four vectors. Or, another interaction is one particle four vector which decays into many other on mass shell four vectors.

This is clear in the pictorial reperentation of Feynman diagrams, which describe the terms of the perturbative expansion the sum of which gives the interaction crossection,( or decay probability.)

Simple example the first order term of the interaction electron scattering off an electron :

e-e-

The famous "collapse" just means one instant in the probability distribution , which is the calculated crossection for the interaction. The quantum mechanical predictions predict the probability of interaction for each event, and a large number of events has to be accumulated in order to test a hypothesis.

but to me EM should make the wave function collapse every instant.

The interactions you envisage are off mass shell, like the photon in the diagram above, they are virtual "interactions" so do not fall into the collapse scenario, which in my opinion is a bad analogy anyway. The wavefunction is not a balloon.

Correct answer by anna v on February 23, 2021

Your question is actually related to what is known as the measurement problem; a yet-unsolved problem in quantum mechanics and the different interpretation of quantum mechanics give different answers.

According to the standard interpretation (Copenhagen interpretation), the wavefunction collapses when interacting with a measurement device not when interacting with other particles. Then the question would be: what does count as a measurement device?

The Copenhagen interpretation does not give a clear definitive answer and simply appeals to the intuitive classical notion of a macroscopic measurement device, thus drawing a vague boundary between the quantum and the classical worlds. This is an unsatisfactory answer because, in the end, any measurement device is actually composed of other elementary particles!

This problem motivated physicists to propose other interpretations of quantum mechanics. Here are a few examples, just to give you an idea of how wildly-different these interpretations are:

  • Spontaneous collapse theory: The wavefunctions of fundamental particles have a fixed probability per time of collapsing. The probability for a single particle is extremely small but a large collection of entangled particles (like in macroscopic objects) would a very high probability.
  • Many-World Interpretation: The Wavefunction never collapses but the world keeps splitting into all the different possible states allowed by the wavefunction.
  • Hidden-variable models like Bohmian mechanics: The particles have definite positions all the time and the wavefunction just represents our ignorance about that.

Answered by Quantum-Collapse on February 23, 2021

to add to what Quantum-Collapse said: interaction between particles (or quantum systems) is manifested in the Hamiltonian operator, and taken into account through the wave function.

for example, coloumb interaction between 2 electrons is represented in the Hamiltonian equation as a potential term, in addition to the kinetic terms of both particles:

$$H=frac{|textbf{p}_1|^2}{2m_e} +frac{|textbf{p}_2|^2}{2m_e}+frac{Ke^2}{|textbf{r}_1-textbf{r}_2|}$$

with the standard definitions of momentum:

$$hat{p}_j=-ihbarnabla_{(j)}$$

in such case you have to solve the Schrodinger equation for a wave function of the 2 positions (or momenta) $psi(textbf{r}_1,textbf{r}_2)$

Answered by Papa Jonathan on February 23, 2021

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